Lecture 17- Bar Development

							Lecture 17- Bar
Development
February 24, 2003
CVEN 444
Lecture Goals
 Slab design reinforcement
 Bar Development
 Hook development
Flexural Reinforcement in
Slabs
For a 1 ft strip of slab is designed like a beam
As (req’d) is in units of (in2/ft)


                       12 in          
  As / ft  Ab                        
                bar spacing in inches 
The table will allow
to determine the
amount of steel per
each foot of slab.
Flexural Reinforcement in
Slabs
The minimum spacing of the bars is given as:
       ACI Sec. 7.6.5
                          3t  slab thickness 
       Smax  smaller of 
                                  18 in.
Also, check crack control - important for exterior
exposure (large cover dimensions) - The spacing S of
reinforcement closest to the surface in tension ACI
Sec. 10.6.4
                540           12  36 
             s      2.5cc 
                 fs              fs
Flexural Reinforcement in
Slabs
Maximum & Minimum reinforcement requirements
  Thin slabs shrink more rapidly than deeper beams.

  Temperature & shrinkage (T&S) steel is provided
  perpendicular to restrain cracks parallel to span.
  (Flexural steel restrains cracks perpendicular to
  span)
Flexural Reinforcement in
Slabs
Maximum & Minimum reinforcement requirements
 T&S Reinforcement (perpendicular to span)
 ACI Sec 7.12.2

As min  0.0020 * 12"* t      f y  40 or 50 ksi
        0.0018* 12"* t         f y  60 ksi
                  60 
        0.0018*   * 12"* t   f y  60 ksi
                 f 
                  y
        0.0014* 12 "* t
Flexural Reinforcement in
Slabs
T&S Reinforcement (perpendicular to span)
ACI Sec 7.12.2.2

                             5t
         S max  smaller of 
                            18"
   t – thickness of the slab
Flexural Reinforcement in
Slabs
Flexural Reinforcement (parallel to span)
ACI Sec 10.5.4
       As min   As min T&S
                              3t
         Smax    smaller of 
                             18"
  Smax from reinforced spacing
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
 A. Concept of Bond Stress and Rebar Anchorage
  Internal Forces in a beam
     Forces developed in the beam by loading.
Reinforcement Development
Lengths, Bar Cutoffs, and
Continuity Requirements
A. Concept of Bond Stress and Rebar Anchorage
  Forces in Rebar
  Bond stresses provide mechanism of force transfer
  between concrete and reinforcement.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
  Equilibrium Condition for Rebar
   F  0  T  Bond Force  0              = bond stress
            d b2
                    f y   d blb   0   (coefficient of
             4                             friction)  k     fc
        ld 
                    f y db
                                                   k  f bar 
                    4
 Note: Bond stress is zero at cracks
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements
Sources of Bond Transfer
  (1) Adhesion between concrete & reinforcement.
  (2) Friction
  Note: These properties are quickly lost for tension.
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements
 (3)Mechanical Interlock.
     The edge stress concentration causes
     cracking to occur.
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements
(3) Mechanical Interlock (cont).
    Force interaction between the steel and concrete.
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements

 Splitting cracks result in loss of bond transfer.
 Reinforcement can be used to restrain these
 cracks.
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements
Splitting Load is Affected by:
  1. Minimum edge distance and spacing of bars
     ( smaller distance = smaller load )
  2. Tensile strength of concrete.
  3. Average bond stress along bar. ( Increase in
     bond stress     larger wedging forces. )
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements
Typical Splitting Failure Surfaces.
Reinforcement Development Lengths,
Bar Cutoffs, and Continuity
Requirements
Typical Splitting Failure Surfaces.
Reinforcement Development Lengths, Bar
Cutoffs, and Continuity Requirements
General splitting of
concrete along the
bars,either in vertical
planes as in figure (a) or
in horizontal plane as in
figure (b). Such splitting
comes largely from
wedging action when the
ribs of the deformed bar
bear against the concrete.
The horizontal type of splitting frequently begins at a diagonal
crack. The dowel action increases the tendency toward splitting.
This indicates that shear and bond failure are often intricately
interrelated.
Reinforcement Development
Lengths
B. ACI Code expression for development length for
   bars in tension/in compression.
   Development Length, ld
    Shortest length of bar in which the bar stress can
    increase from zero to the yield strength, fy.
    ( ld used since bond stresses, , vary along a bar in
    a tension zone)
Reinforcement Development
Lengths
  Development Length, ld

( ld used since bond
stresses, , vary along a
bar in a tension zone)
  Development Length for Bars in
  Tension
  Development length, ld  12” ACI 12.2.1
    fc  10000 psi for Ch. 12 provisions for development length
    in ACI Codes.
  Development length, ld (simplified expression from ACI
  12.2.2)                             No. 6 and smaller No. 7 and larger
                                                  bars and deformed bars
                                                  wires
Clear spacing of bars being developed or
spliced not less than db, clear cover not less
than db, and stirrups or ties throughout ld not    ld   f y      ld   f y
less than the code minimum                                            
                    or                             d b 25 f c       d b 20 f c
Clear spacing of bars being developed or
spliced not less than 2db and clear cover not          38                  47.5
less than db.
Development Length for Bars in
Tension
 Development length, ld  12” ACI 12.2.1
      fc  10000 psi for Ch. 12 provisions for development length in
      ACI Codes.
Development length, ld (simplified expression from ACI 12.2.2)
                                        No. 6 and smaller No. 7 and larger
                                        bars and deformed bars
                                        wires
Other cases                              ld 3 f y        ld 3 f y
                                                              
                                         d b 50 f c         d b 40 f c


                                              57               71

fc = 4 ksi fy = 60 ksi , ,,  1.0
Development Length for Bars in
Tension
Development length, ld    ACI 12.2.3
               ld   3 fy      
                  
               d b 40 f c  c  K ct 
                                    
                           db 
                      c  K ct   
              where,               2.5
                      db         
2.5 limit to safeguard against pullout type failure.
Factors used in expressions for
Development Length (ACI 12.2.4)

  reinforcement location factor       where  < 1.7

 Horizontal reinforcement so placed that
 more than 12 in of fresh concrete is cast      1.3
 in the member below the development
 length or splice
 Other reinforcement                            1.0
Factors used in expressions for
Development Length (ACI 12.2.4)
  coating factor (epoxy prevents adhesion &
                    friction between bar and concrete.)

Epoxy-coated bars or wires with cover less      1.5
than 3db or clear spacing less than 6db
All other epoxy-coated bars or wires            1.2
Uncoated reinforcement                          1.0

              where  < 1.7
Factors used in expressions for
Development Length (ACI 12.2.4)
g  reinforcement size factor (Reflects more favorable
                              performance of smaller 
                              bars)
No.6 and smaller bars and deformed wire       0.8
No. 7 and larger bars                         1.0
Factors used in expressions for
Development Length (ACI 12.2.4)
  lightweight aggregate concrete factor (Reflects
    lower tensile strength of lightweight concrete, &
    resulting reduction in splitting resistance.)

When lightweight aggregate concrete is used.    1.3
However, when fct is specified, shall be        1.0
permitted to be taken as 6.7 f c f ct but not
less than
When normal weight concrete is used             1.0
Factors used in expressions for
Development Length (ACI 12.2.4)
c = spacing or cover dimension, in.
  Use the smaller of either
  (a) the distance from the center of the bar or wire to
  the nearest concrete surface.
  or
  (b) one-half the center-to-center spacing of the bar or
  wires being developed.
Factors used in expressions for
Development Length (ACI 12.2.4)

Ktr = transverse reinforcement index (Represents the
      contribution of confining reinforcement across
      potential splitting planes.)


                       Atr f y t
           K tr 
                    1500 * s * n
Factors used in expressions for
Development Length (ACI 12.2.4)
Atr =   Total cross-section area of all transverse
        reinforcement within the spacing s, which
        crosses the potential plane of splitting along
        the reinforcement being developed with in the
        development length, in2.
fyt =   Specified yield strength of transverse
        reinforcement, psi.
Factors used in expressions for
Development Length (ACI 12.2.4)
   s = maximum center-to-center spacing of
       transverse reinforcement within ld in.
   n = number of bars or wires being developed
       along the plane of splitting.

Note: It is permitted to use Ktr = 0 as a design
      simplification even if transverse reinforcement
      is present.
Excess Flexural Reinforcement
Reduction (ACI 12.2.5)
Reduction = (As req’d ) / (As provided )
   - Except as required for seismic design (see ACI
   21.2.1.4)
   - Good practice to ignore this provision, since use
   of structure may change over time.
   - final ld  12 in.                                Mu
                               M n req'd             
              Reduction                      
                             M n provided       M n provided
Development Length for Bars in
Compression (ACI 12.3)
Compression development length,
ldc = ldbc * applicable reduction factors  8 in.
Basic Development Length for Compression, ldbc
                            0.02 d b f y
                           
        ldbc    larger of       fc
                           0.0003 d b f y
                           
Development Length for Bars in
Compression (ACI 12.3)
 Reduction Factors (ACI 12.3.3)
- Excessive Reinforcement Factor
      = A( s req’d ) / A( s provided)
- Spiral and Ties
       If reinforcement is enclosed with spiral
       reinforcement 0.25 in. diameter and
       4 in.  pitch or within No. 4 ties
       according to 7.10.5 and spaced  4 in.
       on center. Factor = 0.75
Example - Development
For the cross section of a simply
supported beam reinforced with
4 #8 bars that are confined with
#3 stirrup spaced at 6 in.
Determine the development
length of the bars if the beam is
made of normal weight concrete
fc = 3 ksi and fy= 60 ksi
Example - Development
Check if conditions for spacing and concrete
cover are met:
   For #8 bars, db = 1.0 in.
   Clear cover = 2.5 in - 0.5 in.= 2.0 in. > db
   Clear spacing  b  2cover  2d
                                      stirrup
   between bars       spaces
                    12 in.  5 in.
                                   1.0 in.  1.33in.  d b
                          3
Example - Development
Bars are confined with #3 stirrups. The conditions are
met.
      ld  f y
                     (for bars > #7)
      d b 20 f c
Determine the factors; = 1.0 (bottom bars),  =1.0
(no coating) and  = 1.0 (normal weight concrete) and
  fc  54.8 psi < 100 psi
Example - Development
      ld 1.0 1.0 1.0  60000
                                  54.8
      db        20 3000

So ld = 54.8(1.0 in.) = 54.8 in.   55 in. Using the
more general formula Ktr = 0.0
      ld  3   f y  g
                   
      d b  40   f c   c  K tr 
                      
                                   
                          db 
Example - Development
   =  = g =  = 1.0. Also

c = smaller of distance from center of bar to the nearest
    concrete surface c1 or one-half the center-to-center
    of bars spacing c2
                                b  2cover 
                      c2  0.5              
     c1  2.5 in.               spaces 
                                12 in.  5 in. 
                          0.5                   1.17 in.
                                      3        
     c = 1.17 in. controls
Example - Development
    c  K tr 
If            < 1.5 use 1.5. 1.17/1.0 = 1.17   1.5
    db 


  ld  3   60000  1.0
                1.5  54.8
  db  40   3000   

 So ld = 55 in.
Example - Development
If the same beam is made of light
weight aggregate concrete and
the bars are epoxy coated and As
required for analysis is 2.79 in2
Example - Development
The conditions are met.
      ld Rs f y
                         (for bars > #7)
      db   20 f c
Determine the factors; = 1.0 (bottom bars),  =1.5
(epoxy coating) and  = 1.3 (lightweight aggregate
concrete) and Rs = (As(req) / As(provided) ) = 2.79/3.16 = 0.89.
The value of  is 1.5 because the concrete cover is less
than 3db = 3 in.
Example - Development
Check that  =1.0(1.5) = 1.5 < 1.7

      ld 0.89 1.51.3 60000
                               95.1
      db       20 3000

So ld = 95.1(1.0 in.) = 95.1 in.   96 in.
Homework

Problems 10.1 & 10.2